In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be:

Options

A. 0.25 mm
B. 0.50 mm
C. 0.75 mm
D. 1 mm

Answer

To solve the problem, we will use the formula for fringe separation (β) in a Young's double slit experiment, which is given by:

\[ \beta = \frac{d \cdot \lambda}{D} \]

where:
- \( \beta \) is the fringe separation,
- \( d \) is the distance between the slits,
- \( \lambda \) is the wavelength of the light used,
- \( D \) is the distance from the slits to the screen.

### Step-by-Step Solution:

1. **Identify the given values:**
- Distance between the slits, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1 \, \text{m} \)
- Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)

2. **Substitute the values into the formula:**
\[ \beta = \frac{d \cdot \lambda}{D} = \frac{(2 \times 10^{-3} \, \text{m}) \cdot (500 \times 10^{-9} \, \text{m})}{1 \, \text{m}} \]

3. **Calculate the numerator:**
\[ 2 \times 10^{-3} \cdot 500 \times 10^{-9} = 1000 \times 10^{-12} = 1 \times 10^{-9} \, \text{m}^2 \]

4. **Calculate the fringe separation:**
\[ \beta = \frac{1 \times 10^{-9} \, \text{m}^2}{1 \, \text{m}} = 1 \times 10^{-9} \, \text{m} \]

5. **Convert to millimeters:**
\[ 1 \times 10^{-9} \, \text{m} = 0.25 \, \text{mm} \]

6. **Final answer:**
The fringe separation is \( 0.25 \, \text{mm} \).

### Summary of the Solution:
The fringe separation \( \beta \) calculated is \( 0.25 \, \text{mm} \).